path: root/Fundamentals_Of_Thermodynamics/Chapter15_6.ipynb

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  {

   "cells": [

    {

     "cell_type": "heading",

     "level": 1,

     "metadata": {},

     "source": [

      "Chapter15:CHEMICAL REACTIONS"

     ]

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.1:Pg-621"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques1\n",

      "#theoratical air-fuel ratio for combustion of octane\n",

      "#combustion equation is\n",

      "#C8H18 + 12.5O2 + 12.5(3.76) N2 \u2192 8 CO2 + 9H2O + 47.0N2\n",

      "rm=(12.5+47.0)/1;#air fuel ratio on mole basis\n",

      "rma=rm*28.97/114.2;#air fuel ratio on mass basis;\n",

      "print \"Theoratical air fuel ratio on mass basis is\",round(rma),\"kg air/kg fuel\"\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " Theoratical air fuel ratio on mass basis is 15.0 kg air/kg fuel\n"

       ]

      }

     ],

     "prompt_number": 3

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.2:Pg-622"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 2\n",

      "# molal analysis of the products of combustion\n",

      "# combustion equation is \n",

      "# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 \u2192 8 CO2 + 9H2O + 12.5O2 + 94.0N2\n",

      "nCO2=8 # number of moles of CO2 \n",

      "nH2O=9 # number of moles of H2O\n",

      "nO2=12.5 # number of moles of O2\n",

      "nN2=94.0 # number of moles of N2\n",

      "moles=nCO2+nH2O+nO2+nN2 # total moles\n",

      "# molal analysis of products\n",

      "xCO2=nCO2*100/moles \n",

      "xH2O=nH2O*100/moles\n",

      "xO2=nO2*100/moles\n",

      "xN2=nN2*100/moles\n",

      "print \" CO2 =\",round(xCO2,2)\n",

      "print \"\\n H2O=\",round(xH2O,2)\n",

      "print \"\\n O2=\",round(xO2,2)\n",

      "print \"\\n N2=\",round(xN2,2)\n",

      "print \"\\n The dew point temp. corresponding to partial pressure of water i.e \",round(xH2O,2),\"KPa is 39.7 degree C\""

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " CO2 = 6.48\n",

        "\n",

        " H2O= 7.29\n",

        "\n",

        " O2= 10.12\n",

        "\n",

        " N2= 76.11\n",

        "\n",

        " The dew point temp. corresponding to partial pressure of water i.e  7.29 KPa is 39.7 degree C\n"

       ]

      }

     ],

     "prompt_number": 6

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.2E:Pg-622"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 2\n",

      "# molal analysis of the products of combustion\n",

      "# combustion equation is \n",

      "# C8H18 + 12.5(2) O2 + 12.5(2)(3.76) N2 \u2192 8 CO2 + 9H2O + 12.5O2 + 94.0N2\n",

      "pressure=14.7 # in lbf/in^2\n",

      "nCO2=8 # number of moles of CO2 \n",

      "nH2O=9 # number of moles of H2O\n",

      "nO2=12.5 # number of moles of O2\n",

      "nN2=94.0 # number of moles of N2\n",

      "moles=nCO2+nH2O+nO2+nN2 # total moles\n",

      "# molal analysis of products\n",

      "xCO2=nCO2*100/moles \n",

      "xH2O=nH2O*100/moles\n",

      "xO2=nO2*100/moles\n",

      "xN2=nN2*100/moles\n",

      "print \" CO2 =\",round(xCO2,2)\n",

      "print \"\\n H2O=\",round(xH2O,2)\n",

      "print \"\\n O2=\",round(xO2,2)\n",

      "print \"\\n N2=\",round(xN2,2)\n",

      "PressureO2=pressure*xH2O/100 # partial pressure of O2\n",

      "print \"\\n The dew point temp. corresponding to partial pressure of water i.e \",round(PressureO2,3),\"lbf/in^2 is 104 F degree C\"\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " CO2 = 6.48\n",

        "\n",

        " H2O= 7.29\n",

        "\n",

        " O2= 10.12\n",

        "\n",

        " N2= 76.11\n",

        "\n",

        " The dew point temp. corresponding to partial pressure of water i.e  1.071 lbf/in^2 is 104 F degree C\n"

       ]

      }

     ],

     "prompt_number": 8

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.3:Pg-623"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 3\n",

      "# to determine the air\u2013fuel ratio on a volumetric basis\n",

      "# the combustion equation is\n",

      "# 0.14H2+0.27CO+0.03CH4+0.006O2+0.509N2+0.045CO2+0.259O2+0.259(3.76)N2\u21920.20H2O+0.345CO2+1.482 N2\n",

      "mH2=2 # molar mass of H2\n",

      "mCO=28 # molar mass of CO\n",

      "mCH4=16 # molar mass of CH4\n",

      "mO2=32 # molar mass of O2\n",

      "mN2=28 # molar mass of N2\n",

      "mCO2=44 # molar mass of CO2\n",

      "\n",

      "\n",

      "kmolair=0.259 # moles of air in mixture using combustion equation\n",

      "kmolfuel=0.21 # moles of fuel in mixture using combustion equation\n",

      "ratioth=kmolair/kmolfuel # theoratical ratio\n",

      "\n",

      "# For 20% excess air\n",

      "vratio=ratioth*(1+0.2) # new ratio for excess air on volume basis\n",

      "\n",

      "print \" The A/F ratio on volumetric basis is \",round(vratio,2)\n",

      "mratio=vratio*28.97 /(0.14*mH2+0.27*mCO+0.03*mCH4+0.006*mO2+0.509*mN2+0.045*mCO2)\n",

      "print \"\\n The A/F ratio on mass basis is \",round(mratio,2),\"kg air/kg fuel\"\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The A/F ratio on volumetric basis is  1.48\n",

        "\n",

        " The A/F ratio on mass basis is  1.73 kg air/kg fuel\n"

       ]

      }

     ],

     "prompt_number": 11

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.4:Pg-624"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 4\n",

      "# determining the air\u2013fuel ratio and the percent theoretical air\n",

      "CO2=10.00 # percentage\n",

      "O2=2.37 # percentage\n",

      "CO=0.53 # percentage\n",

      "N2=87.10 # percentage\n",

      "mCH4=16 # molecular mass of CH4\n",

      "# using this data the combustion equation is\n",

      "# a CH4 + b O2 + c N2 \u2192 10.0 CO2 + 0.53 CO + 2.37 O2 + d H2O + 87.1N2\n",

      "# on balancing we get\n",

      "# 10.53 CH4 + 23.16 O2 + 87.1N2 \u219210.0 CO2 + 0.53 CO + 2.37 O2 + 21.06 H2O + 87.1N2\n",

      "# balanced combustion equation for per Kmole of fuel is:\n",

      "# CH4 + 2.2O2 + 8.27 N2 \u2192 0.95 CO2 + 0.05 CO + 2H2O + 0.225 O2 + 8.27 N2\n",

      "molesAir=2.2+8.27 # moles of air\n",

      "AF= 28.97*molesAir/mCH4 # air fuel ratio on mass basis\n",

      "print \"\\n The A/F ratio on mass basis is \",round(AF,2),\"kg air/kg fuel\"\n",

      "# Theoritical combustion equation:\n",

      "# CH4 + 2O2 + 2(3.76)N2 \u2192 CO2 + 2H2O + 7.52 N2\n",

      "molesAirth=2+2*3.76 # moles of air(theoritical)\n",

      "AFth= 28.97*molesAirth/mCH4 # air fuel ratio on mass basis (theoritical)\n",

      "print \"\\n The A/F ratio (theoritical) on mass basis is \",round(AFth,2),\"kg air/kg fuel\"\n",

      "print \"\\n The percent theoretical air is\",round(AF*100/AFth,1)"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " The A/F ratio on mass basis is  18.96 kg air/kg fuel\n",

        "\n",

        " The A/F ratio (theoritical) on mass basis is  17.24 kg air/kg fuel\n",

        "\n",

        " The percent theoretical air is 110.0\n"

       ]

      }

     ],

     "prompt_number": 16

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.5:Pg-625"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 5\n",

      "\n",

      "Sulfur=0.6 # percent by mass\n",

      "Hydrogen=5.7 # percent by mass\n",

      "Carbon=79.2 # percent by mass\n",

      "Oxygen=10.0 # percent by mass\n",

      "Nitrogen=1.5 # percent by mass\n",

      "Ash=3.0 # percent by mass\n",

      "mS=32.0 # molecular mass \n",

      "mH2=2.0  # molecular mass\n",

      "mC=12.0  # molecular mass\n",

      "mO2=32.0  # molecular mass\n",

      "mN2=28.0  # molecular mass\n",

      "# molal composition per 100 Kg of fuel:\n",

      "S=Sulfur/mS # molal composition \n",

      "H=Hydrogen/mH2 # molal composition\n",

      "C=Carbon/mC # molal composition\n",

      "O2=Oxygen/mO2 # molal composition\n",

      "N2=Nitrogen/mN2 # molal composition\n",

      "\n",

      "#The combustion equation becomes \n",

      "# 0.02 S + 0.02 O2 \u21920.02 SO2\n",

      "# 2.85 H2 + 1.42 O2 \u21922.85 H2O\n",

      "# 6.60 C + 6.60 O2 \u21926.60 CO2\n",

      "\n",

      "O2req=0.02+1.42+6.60 # kmol O2 required/100 kg fuel\n",

      "O2present=O2req-O2 # kmol O2 from air/100 kg fuel\n",

      "AFtheo= (O2present+O2present*3.76)*28.97/100 # AF ration theoritical kg air/kg fuel\n",

      "print \"\\n The A/F ratio (theoritical) on mass basis is \",round(AFtheo,2),\"kg air/kg fuel\"\n",

      "print \"\\n The A/F ratio for 30% excess air on mass basis is \",round(AFtheo*1.3,2),\"kg air/kg fuel\"\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "\n",

        " The A/F ratio (theoritical) on mass basis is  10.66 kg air/kg fuel\n",

        "\n",

        " The A/F ratio for 30% excess air on mass basis is  13.85 kg air/kg fuel\n"

       ]

      }

     ],

     "prompt_number": 23

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.6:Pg-629"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques6\n",

      "# determining heat transfer per kilomole of fuel entering combustion chamber\n",

      "\n",

      "#1-CH4\n",

      "#2-CO2\n",

      "#3-H2O\n",

      "#hf-standard enthalpy of given substance\n",

      "hf1=-74.873;#kJ\n",

      "hf2=-393.522;#kJ\n",

      "hf3=-285.830;#kJ\n",

      "Qcv=hf2+2*hf3-hf1;#kJ\n",

      "print \"Heat transfer per kilomole of fuel entering combustion chamber is\",round(Qcv,3),\"kJ\"\n",

      "#the answers in the book is different as they have not printed the decimals in values"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "Heat transfer per kilomole of fuel entering combustion chamber is -890.309 kJ\n"

       ]

      }

     ],

     "prompt_number": 5

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.7:Pg-631"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques7\n",

      "#calculating enthalpy of water at given pressure and temperature\n",

      "\n",

      "#1.Assuming steam to be an ideal gas with value of Cp\n",

      "T1=298.15;#Initial temperature in K\n",

      "T2=573.15;#final temperature in K\n",

      "T=(T1+T2)/2;#average temperature in K\n",

      "Cp=1.79+0.107*T/1000+0.586*(T/1000)**2-.20*(T/1000)**3;#specific heat at constant pressure in kj/kg.K\n",

      "M=18.015;#mass in kg\n",

      "dh=M*Cp*(T2-T1);#enthalpy change in kJ/kmol\n",

      "ho=-241.826;#enthalpy at standard temperature and pressure in kJ/mol\n",

      "htp1=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

      "print \" 1. Enthalpy of water at given pressure and temperature using value of Cp =\",round(htp1,3),\"kJ/kmol\"\n",

      "\n",

      "#2..Assuming steam to be an ideal gas with value for dh\n",

      "dh=9359;#enthalpy change from table A.9 in kJ/mol\n",

      "htp2=ho+dh/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

      "print \" 2. Enthalpy of water at given pressure and temperature assuming value od dh =\",round(htp2,3),\"kJ/kmol \"\n",

      "\n",

      "#3. Using steam table\n",

      "dh=M*(2977.5-2547.2);#enthalpy change for gases in kJ/mol\n",

      "htp3g=dh/1000+ho;\n",

      "dh=M*(2977.5-104.9);#enthalpy change for liquid in kJ/mol\n",

      "hl=-285.830;#standard enthalpy for liquid in kJ/kmol\n",

      "htp31=hl+dh/1000.0;#enthalpy at given temp and pressure in kJ/kmol\n",

      "print \" 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp31,3),\"kJ/kmol \"\n",

      "print \" 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid =\",round(htp3g,3),\"kJ/kmol \"\n",

      "#4.using generalised charts\n",

      "#htp=ho-(h2*-h2)+(h2*-h1*)+(h1*-h1);\n",

      "#h2*-h2=Z*R*Tc,\n",

      "#h2*-h1*=9539 kJ/mol, from part 2\n",

      "#h1*-h1=0 ,as ideal gas \n",

      "Z=0.21;#from chart\n",

      "R=8.3145;#gas constant in SI units\n",

      "Tc=647.3;#critical temperature in K\n",

      "htp4=ho+9539/1000-Z*R*Tc/1000;#enthalpy at given temp and pressure in kJ/kmol\n",

      "print \" 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart = \",round(htp4,3),\"kJ/kmol\"\n",

      "#the answers in book are different as they have not printed the decimals in values"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " 1. Enthalpy of water at given pressure and temperature using value of Cp = -232.258 kJ/kmol\n",

        " 2. Enthalpy of water at given pressure and temperature assuming value od dh = -232.826 kJ/kmol \n",

        " 3.(i) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.08 kJ/kmol \n",

        " 3.(ii) enthalpy at given temp and pressure in kJ/kmol in terms of liquid = -234.074 kJ/kmol \n",

        " 4. enthalpy at given temp and pressure in kJ/kmol using compressibility chart =  -233.956 kJ/kmol\n"

       ]

      }

     ],

     "prompt_number": 12

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.8:Pg-632"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 8\n",

      "# Determine the heat transfer\n",

      "airth=400 # % excess theoritical air\n",

      "Ti=25 # Entry temp in degree celsius\n",

      "To=900 # Exit temp in Kelvin\n",

      "mf=0.25 # specific fuel consumption in kg/s of fuel per megawatt output\n",

      "# The combustion equation is:\n",

      "# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 \u2192 8CO2 + 9H2O + 37.5O2 + 188.0N2\n",

      "hfC8H18=-250105 # is the enthalpy of fuel in kJ/kmol \n",

      "nCO2=8 # moles\n",

      "hfCO2=-393522 # enthalpy in kJ/kmol\n",

      "DelhCO2=28030 # enthalpy in kJ/kmol\n",

      "mC8H18=114.23 # molecular mass\n",

      "nH2O=9 # moles\n",

      "hfH2O=-241826 # enthalpy in kJ/kmol\n",

      "DelhH2O=21937 # enthalpy in kJ/kmol\n",

      "DelhO2=19241 # enthalpy in kJ/kmol\n",

      "nO2=37.5 # moles\n",

      "DelhN2=18225 # enthalpy in kJ/kmol\n",

      "nN2=188.0 # moles\n",

      "Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2\n",

      "\n",

      "Wcv=1000*mC8H18/mf # work in kJ/kmol fuel\n",

      "print \" The work is \",Wcv,\"kJ/kmol fuel\"\n",

      "\n",

      "Qcv=Eproduct+Wcv-hfC8H18 # using first law\n",

      "print \" The Heat transfer is \",Qcv,\"kJ/kmol fuel\"\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The work is  456920.0 kJ/kmol fuel\n",

        " The Heat transfer is  -48074.5 kJ/kmol fuel\n"

       ]

      }

     ],

     "prompt_number": 27

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.8E:Pg-633"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 8\n",

      "# Determine the heat transfer\n",

      "airth=400 # % excess theoritical air\n",

      "Ti=77 # Entry temp in F\n",

      "To=1100 # Exit temp in F\n",

      "mf=1 # specific fuel consumption in lb of fuel per horsepower-hour\n",

      "# The combustion equation is:\n",

      "# C8H18(l) + 4(12.5)O2 + 4(12.5)(3.76)N2 \u2192 8CO2 + 9H2O + 37.5O2 + 188.0N2\n",

      "hfC8H18=-107526 # is the enthalpy of fuel in Btu/lb.mol\n",

      "nCO2=8 # moles\n",

      "hfCO2=-169184 # enthalpy in Btu/lb.mol\n",

      "DelhCO2=11391 # enthalpy in Btu/lb.mol\n",

      "mC8H18=114.23 # molecular mass\n",

      "nH2O=9 # moles\n",

      "hfH2O=-103966 # enthalpy in Btu/lb.mol\n",

      "DelhH2O=8867 # enthalpy in Btu/lb.mol\n",

      "DelhO2=7784 # enthalpy in Btu/lb.mol\n",

      "nO2=37.5 # moles\n",

      "DelhN2=7374 # enthalpy in Btu/lb.mol\n",

      "nN2=188.0 # moles\n",

      "Eproduct=nCO2*(hfCO2+DelhCO2)+nH2O*(hfH2O+DelhH2O)+nO2*DelhO2+nN2*DelhN2\n",

      "\n",

      "Wcv=2544*mC8H18 # work in kJ/kmol fuel\n",

      "print \" The work is \",Wcv,\"Btu/lb.mol fuel\"\n",

      "\n",

      "Qcv=Eproduct+Wcv-hfC8H18 # using first law\n",

      "print \" The Heat transfer is \",Qcv,\"Btu/lb.mol fuel\"\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The work is  290601.12 Btu/lb.mol fuel\n",

        " The Heat transfer is  -41895.88 Btu/lb.mol fuel\n"

       ]

      }

     ],

     "prompt_number": 31

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.9:Pg-634"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques 9\n",

      "\n",

      "# The combustion equation is\n",

      "# C2H4 + 3O2 \u2192 2 CO2 + 2H2O(g)\n",

      "hfC2H4=52467 # enthalpy in KJ/mol\n",

      "R=8.314 # gas constant\n",

      "T=298.2 # temperature in Kelvin\n",

      "hfCO2=-393522 # enthalpy in kJ/kmol\n",

      "DelhCO2=12906 # enthalpy in kJ/kmol\n",

      "hfH2O=-241826 # enthalpy in kJ/kmol\n",

      "DelhH2O=10499 # enthalpy in kJ/kmol\n",

      "Ureactant=hfC2H4-4*R*T # energy of reactant in KJ\n",

      "Uproduct=2*(hfCO2+DelhCO2)+2*(hfH2O+DelhH2O) # energy of product in KJ\n",

      "Q=Uproduct-Ureactant # using first law\n",

      "print \"The Heat transfer is \",Q,\"kJ\"\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The Heat transfer is  -1266436.0608 kJ\n"

       ]

      }

     ],

     "prompt_number": 33

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.10:Pg-638"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques 10\n",

      "\n",

      "# The combustion equation is\n",

      "# C3H8 + 5O2 \u2192 3 CO2 + 4H2O\n",

      "hfC3H8g=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10\n",

      "hfC3H8l=hfC3H8g-44.097*370 # enthalpy of C3H8 liquid as enthalpy of evaporation is given in question\n",

      "hfH2Ol=285830 # enthalpy of water in KJ/mol\n",

      "hfH2Og=241826 # enthalpy in KJ/mol\n",

      "hfCO2=393522 # enthalpy in KJ/mol\n",

      "nO2=5 # moles\n",

      "nCO2=3 # moles\n",

      "nH2O=4 # moles\n",

      "nC3H8=1 # moles\n",

      "# part(1) : liquid propane-liquid water\n",

      "hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8l)/44.097 # higher heating value of liquid propane \n",

      "print \"The higher heating value of liquid propane is\",round(hrP,2),\"KJ/kg\"\n",

      "\n",

      "# part(2) : Liquid propane\u2013gaseous water\n",

      "\n",

      "hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8l)/44.097 # lower heating value of liquid propane \n",

      "print \"The lower heating value of liquid propane is\",round(hrP,2),\"KJ/kg\"\n",

      "\n",

      "# part(3) :Gaseous propane\u2013liquid water\n",

      "hrP=(nCO2*hfCO2 + nH2O*hfH2Ol -nC3H8*hfC3H8g)/44.097 # higher heating value of gaseous propane \n",

      "print \"The higher heating value of gaseous propane is\",round(hrP,2),\"KJ/kg\"\n",

      "\n",

      "# part(4) :Gaseous propane\u2013Gaseous water\n",

      "hrP=(nCO2*hfCO2 + nH2O*hfH2Og -nC3H8*hfC3H8g)/44.097 # lower heating value of gaseous propane \n",

      "print \"The lower heating value of gaseous propane is\",round(hrP,2),\"KJ/kg\"\n",

      "\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n",

      "\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The higher heating value of liquid propane is -120215.89 KJ/kg\n",

        "The lower heating value of liquid propane is 51434.02 KJ/kg\n",

        "The higher heating value of gaseous propane is 55055.58 KJ/kg\n",

        "The lower heating value of gaseous propane is 51064.02 KJ/kg\n"

       ]

      }

     ],

     "prompt_number": 3

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.11:Pg-638"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques 10\n",

      "# The combustion equation is\n",

      "# C3H8(g) + 5O2 \u2192 3 CO2 + 4H2O(g)\n",

      "hfC3H8=-103900 # enthalpy of C3H8 in KJ/mol from Table A.10\n",

      "Cp=2.1 # in KJ/Kg.K\n",

      "T1=500 # temp in Kelvin\n",

      "T2=298.2 # temp in Kelvin\n",

      "DelhO2=6086 # enthalpy of O2 in KJ/mol\n",

      "hfCO2=-393522 # enthalpy in KJ/mol\n",

      "DelhCO2=8305 # enthalpy of CO2 in KJ/mol\n",

      "hfH2O=-241826 # enthalpy of H2O in KJ/mol\n",

      "DelhH2O=6922 # enthalpy in KJ/mol\n",

      "nO2=5 # moles\n",

      "nCO2=3 # moles\n",

      "nH2O=4 # moles\n",

      "nC3H8=1 # moles\n",

      "\n",

      "hR500= hfC3H8+Cp*44.097*(T1-T2)+nO2*DelhO2  # in KJ/mol\n",

      "hP500= nCO2*(hfCO2+DelhCO2)+ nH2O*(DelhH2O+hfH2O) # in KJ/mol\n",

      "hRP500=hP500-hR500 #  in KJ/mol\n",

      "Hrp500=hRP500/44.097 # KJ/Kg\n",

      "\n",

      "print \" The enthalpy of combustion of gaseous propane at 500 K \",round(Hrp500)\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The enthalpy of combustion of gaseous propane at 500 K  -46273.0\n"

       ]

      }

     ],

     "prompt_number": 10

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.12:Pg-639"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 12\n",

      "# The question is solved by Trial and Error and  using the table , no need for Calculation Thus not solved in Python"

     ],

     "language": "python",

     "metadata": {},

     "outputs": []

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.13:Pg-645"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 13\n",

      "# The equation for this chemical reaction is\n",

      "# C2H4(g) + 3(4) O2 + 3(4)(3.76) N2 \u2192 2 CO2+2H2O(g) + 9O2 + 45.1 N2\n",

      "T=25+273.15 # temp in degree kelvin\n",

      "P=0.1 # pressure in Mpa\n",

      "hfCO2=-393522 # enthalpy in KJ/mol\n",

      "hfH2O=-241826 # enthalpy of H2O in KJ/mol\n",

      "HfC2H4=52467 # enthalpy of C2H4 in KJ/mol\n",

      "hfO2=0 # enthalpy of formation in KJ/mol\n",

      "SCO2=213.795 # kJ/kmol fuel\n",

      "SH2O=188.243 # kJ/kmol fuel\n",

      "SC2H4=219.330 # kJ/kmol fuel\n",

      "SO2=205.148 # kJ/kmol fuel\n",

      "DelH=2*hfCO2+2*hfH2O-HfC2H4-3*hfO2 # in KJ/mol\n",

      "DelS=2*SCO2+2*SH2O-SC2H4-3*SO2 # kJ/kmol fuel\n",

      "DelG=DelH-T*(DelS) # Gibbs Energy\n",

      "Wrev=-DelG/28.054 # reversible work in kJ/kg\n",

      "print \"The reversible work is \",round(Wrev,2),\"kJ/kg \"\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        "The reversible work is  46838.61 kJ/kg \n"

       ]

      }

     ],

     "prompt_number": 12

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.14:Pg-647"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques \n",

      "# The question is solved by Trial and Error and  using the table , no need for Calculation Thus not solved in Python"

     ],

     "language": "python",

     "metadata": {},

     "outputs": []

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.15:Pg-649"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques15\n",

      "#calculatng reversible elecromotive force \n",

      "\n",

      "#1-H2O\n",

      "#2-H2\n",

      "#3-O2\n",

      "#hf-standard enthalpy \n",

      "#sf-standard entropy\n",

      "hf1=-285.830;#kJ\n",

      "hf2=0;#kJ\n",

      "hf3=0;#kJ\n",

      "sf1=69.950;#kJ/K\n",

      "sf2=130.678;#kJ/K\n",

      "sf3=205.148;#kJ/K\n",

      "dH=2*hf1-2*hf2-hf3;#change in enthalpy in kJ\n",

      "dS=2*sf1-2*sf2-sf3;#change in entropy in kJ/K\n",

      "T=298.15;#temperature in K\n",

      "dG=dH-T*dS/1000;#change in gibbs free energy in kJ\n",

      "E=-dG*1000/(96485*4);#emf in V\n",

      "print\" Reversible electromotive Force =\",round(E,3),\" V\"\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " Reversible electromotive Force = 1.229  V\n"

       ]

      }

     ],

     "prompt_number": 15

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.16:Pg-654"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "# ques 16\n",

      "# to Calculate the combustion efficiency\n",

      "Tair = 400  # temp in K\n",

      "Vair = 100 # velocity in m/s\n",

      "Tfuel = 50 # temp of fuel in \u25e6C\n",

      "Tproducts = 1100 # temp of product in K\n",

      "Vproducts = 150 # velocity of product in m/s\n",

      "FAactual = 0.0211 # air fuel ratio kg fuel/kg air\n",

      "\n",

      "# HR + KER= \u2212243737 + 14892nO2 \n",

      "# HP + KEP= \u22125 068 599 + 120 853nO2 \n",

      "# using these two equations :\n",

      "no2=(-5068599.0 + 243737.0)/(14892.0-120853) # in kmol O2/kmol fuel\n",

      "FAideal=114.23/(4.76*no2*28.91) # in kg fuel/kg air\n",

      "\n",

      "efficomb=FAideal*100/FAactual # Efficiency\n",

      "\n",

      "print \" The combustion efficiency is \",round(efficomb,2),\"%\"\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise\n"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " The combustion efficiency is  86.4 %\n"

       ]

      }

     ],

     "prompt_number": 16

    },

    {

     "cell_type": "heading",

     "level": 2,

     "metadata": {},

     "source": [

      "Ex15.17:Pg-654"

     ]

    },

    {

     "cell_type": "code",

     "collapsed": false,

     "input": [

      "#ques17\n",

      "#efficiency of generator and plant\n",

      "\n",

      "q=325000*(3398.3-856.0);#heat transferred to H2O/kg fuel in kJ/kg\n",

      "qv=26700.0*33250;#higher heating value in kJ/kg\n",

      "nst=q/qv*100;#efficiency of steam generator\n",

      "w=81000.0*3600;#net work done in kJ/kg\n",

      "nth=w/qv*100.0;#thermal efficiency\n",

      "print\" Efficiency of generator =\",round(nst,1),\"percent\\n\"\n",

      "print\" Thermal Efficiency =\",round(nth,1),\" percent\"\n",

      "\n",

      "# the answer is slightly different in textbook due to approximation while here the answers are precise"

     ],

     "language": "python",

     "metadata": {},

     "outputs": [

      {

       "output_type": "stream",

       "stream": "stdout",

       "text": [

        " Efficiency of generator = 93.1 percent\n",

        "\n",

        " Thermal Efficiency = 32.8  percent\n"

       ]

      }

     ],

     "prompt_number": 13

    }

   ],

   "metadata": {}

  }

 ]

}